Question

Determine if the Mean Value Theorem for Integrals applies to the function f(x) = x3 − 9x on the interval [−1, 1]. If so, find the x-coordinates of the point(s) guaranteed to exist by the theorem

Answer 1

Is the function continuous on the interval?

Is the interval closed?

Answer 2

um it is continuous and i do believe it is closed

Answer 3

Ok, so the theorem applies.

Answer 4

oh really, okay great that was actually the part i was most unsure of

Answer 5

Now we need to look at

\[F(b) - F(a)=F'(c)(b-a) \] and we have \(F(x) = \int_a^xf(t) dt\) where \(a=-1\) and \(b=1\).

Answer 6

okay so then
\[ \int\limits_{-1}^{1}x^3 -9x\]

Answer 7

so is c the x-coordinate im looking for?

Answer 8

\(\int_{-1}^1(t^3-9t) dt = 2(c^3-9c)\) Correct.

Answer 9

okay so im getting 2/3 fro F(b) - F(a)

Answer 10

do i just solve for c now?

Answer 11

hmm cubic around the origin shuold even out on that interval

Answer 12

http://www.wolframalpha.com/input/?i=%5Cint_%7B-1%7D%5E1%28t%5E3-9t%29+dt

Answer 13

but yes, solve for \(c\). I get

\(0=2c(c^2-9)\)

Answer 14

oh woops i see why i got 2/3 , i put t^2 instead of t^3

Answer 15

\(-3\) is not in our interval.

Answer 16

are you sure i should leave it as zero and not 8.5

Answer 17

?

Answer 18

okay nvm sorry iv been solving a lot of area and volume problems like these in which area and volume cant be negative so i am told to interpret negatives as positives put i guess it doesn't apply here :)

Answer 19

couldn't c = 0

Answer 20

\[0 = 2c(c^2-9)\]
\[0 = 2c\]
\[0 = c\]

Answer 21

\(0=2c(c^2-9)\\\dfrac{0}{2c}=c^2-9\\0=c^2-9\\c^2=9\)

Answer 22

c = 3 , 0

Answer 23

correct

Answer 24

so is my answer 0 because 3 is not in the given interval?

Answer 25

yes ...sorry

Answer 26

okay great, thanks for the help :D

Answer 27

I think at some point i said \(3\in [-1,1]\) doooh. deleted it...

Answer 28

np m8